Interpretación de las imágenes

Moving Source, Stationary Observer on Line of Motion

The source can be considered to emit wave pulses at regular intervals and waves are emitted in both the positive and the negative x-direction. The slopes of the wave lines are determined only by the wave speed c in the surrounding air, and with the speed u of the source smaller than the wave speed, the slope of each line is smaller than the slope of the source trajectory.

From the wave lines, we get an idea of the time dependence of the wave trains recorded by observers at rest ahead of and behind the source. It is clear that the number of wave lines (wave pulses or periods) observed per second ahead of the source, at x= x1, will be greater than behind it, at x = x2. In the case of harmonic time dependence, the wave lines can be thought of as representing the crests of the waves. In that case the number of lines per second will be the observed frequency of the harmonic wave.

These frequencies can be obtained in several ways. Since the source moves with a velocity u in the x-direction as it emits a harmonic wave, the separation of the wave maxima ‘imprinted’ on the gas and constituting the emitted sound wave will not be the ordinary wavelength. With reference to Fig. 3.1, consider one wave front emitted in

Figure 3.1: Doppler effect. Moving source, stationary observer. Source velocity: u. Sound speed: c. Wave fronts are shown as vertical lines.

the direction of motion at x= 0 and t = 0 and another at t = T , one period later. The first wave front will be at x= cT when the other wave front is emitted, thus located at x = 0. The source has then reached the position ut. This means that the separation of the imprinted wave fronts will be λ^= (c−u)T which is the wavelength of the wave that travels with the wave speed c. It is shorter than the wavelength λ= c/T which would have been obtained if the source had been at rest. For the sound traveling in the opposite direction, the wavelength will be λ^= (c + u)T .

The wavefronts come closer together in the forward direction and further apart in the opposite direction. The frequency of the emitted wave from the moving source, as observed by a stationary observer ahead of the source, will be f1= c/λ^= c/[(c − u)T] = f/(1 − m), where f is the frequency of the source and m = u/c, the Mach number of the source. The corresponding observed frequency f2 for an observer behind the source is obtained by merely changing the sign of m. Consequently,

f₁= f/(1 − m)

f₂= f/(1 + m), (3.10)

where m = u/c. These relations express the Doppler effect. The difference in frequency f1− f (or f − f2) is referred to as the Doppler shift. It is important to understand that u is the speed of the source relative to the observer. If the absorber is not located on the line of motion of the source, it is the velocity component of the source in the direction of the observer which counts. Thus, when the sound emitted at an angle φ with respect to the direction of motion of the source, the Doppler shift in this direction is determined by the velocity component u cos φ so that m in Eq. 3.10 should be replaced by m cos φ. It is important to realize, however, that when the sound arrives at the observer, the source has moved so that the emission angle is not the same as the view angle under under which the source is seen at the time of arrival of the Doppler shifted sound. This is explained further in the example given below.

The Doppler effect occurs for all waves. The frequency of the light from a source moving away from us is down shifted (toward the red part of the spectrum) and the shift is usually referred to as the ‘red-shift.’

Another way to obtain Eq. 3.10 is geometrical, using a wave diagram. This is done in Example 20 in Chapter 11. The diagram used there looks a bit complicated because of the many lines involved; perhaps you can simplify it.

Eq. 3.10 is valid when the source speed is smaller than the wave speed, i.e., when m <1. For supersonic motion of the source, m > 1, we get f1 = f/(m − 1) and f₂= f/(m + 1).

The Doppler effect is important throughout physics. It is used in a wide range of applications both technical and scientific for the measurement of the speed of moving objects ranging from molecules to galaxies.

In the case of sound from a source like an aircraft, the speed of the source can exceed the wave speed. The slope of the trajectory of the source will then be smaller than the slope of the wave lines, and it follows that the wave lines emitted in the positive and negative x-directions will emerge on the same side of the source trajectory and cross each other; this indicates interference between forward and backward running wave.

Observer on Side Line

The observer is now located at a distance h from the line of motion of the source.

At time t, the location of the source is at xs = ut. The wave reaching the observer at this time was emitted at an earlier time tefrom the emission point xs = ute. The distance from this emission point to the observer can be expressed as R= c(t − te= c(x− xe)/u. With the coordinates of the observation point being x, y, R can be calculated from R²= y²+(x−xe)². With x−xe= x−xs+(xs−xe)= x−ut +uR/v the equation for R can be written R²= y²+ [x − ut + u(R/v)]²with the solution

R = [m(x − ut) ± R1/(1− m²)

R₁= [(x − ut)²+ (1 − m²)y²]^1/2, (3.11) where m= u/c is the Mach number of the source. The distance R must be positive, and for subsonic motion only the plus sign corresponds to a physically acceptable solution.

With the emission angle between the line of propagation from the emission to the observation point denoted φ, the component of the source velocity in this direction will be u cos φ. The Doppler shifted frequency depends only on this component and is f^= f/(1−m cos φ). This Doppler shifted frequency can be expressed in terms of the observer coordinates and time and we leave it for one of the problems to show that f^= f/(1 = m cos φ) = f (R/R1), (3.12) where R and R1are given in Eq. 9.29.

For large negative values of the source location xs, the component of the source velocity in the direction of the observation point is approximately u, and the corre-sponding Doppler shifted frequency is the f/(1− m). Similarly, after the source has passed the observer, the frequency approaches the value f/(1+ m) asymptotically.

For example, with a source Mach number of 0.9 the corresponding range in Doppler shifted frequencies goes from 10 f to 0.53 f .

Although the Doppler shift is zero when the emission angle is 90 degrees, there is an upshift in frequency when the source is at xs = 0. The reason is that xs = 0 corresponds to an emission point at an earlier time and the emission angle is less than 90 degrees. For a source with supersonic speed there are two emission points that contribute to the sound pressure at time t, as illustrated in Fig. 3.2. The corresponding travel distances R^and R^correspond to the two solutions in Eq. 9.29 in which now

Figure 3.2: For a supersonically moving source, there are two emission points contributing to the sound pressure at the stationary observer O.

both the plus and minus signs are acceptable. There are corresponding emission angles φ^and φ^and the Doppler shifted frequency for each can be calculated from Eq. 3.12. It should be noted though, that f/(1− m cos φ^)becomes negative. This merely means that the wave fronts emitted from the source arrive in reverse order, the front emitted last arrives first.

Stationary Source, Moving Observer

Consider two successive wave fronts emitted from the stationary source S separated in time by T and in space by λ. These wave fronts travel with the velocity c, the sound speed. The observer O is moving with the velocity u0. The time it takes for these front to pass the observer is then T^= λ/(c − u0)and the corresponding frequency, f^= 1/T^is

f^= f (1 − m0), (3.13)

where m0= u0/cis the Mach number of the observer.

It is instructive to consider the waveline interpretation of this case.

Both Source and Observer Moving

As before, denote by u and u0the velocities of the source and the observer along the x-axis. The distance between two wave fronts emitted a time T apart will be (c−u)T , where c is the sound speed. The two wave fronts travel with the speed c− u0with respect to the observer. The time required for the wave fronts to pass by will be T^= (c − u)T /(c − u0)and the corresponding observed frequency

f^= f (1 − m0)/(1− m), (3.14)

where, as before, m and m0are the Mach numbers of the source and receiver. For small values of m and m0, we get f^ ≈ f [1 + (m − m0)] which depends only on the relative velocity of the source and the observer. For electromagnetic waves in vacuum, the Doppler shift depends only on the relative speed under all conditions.

This is a consequence of the speed of light being the same in all frames of reference.

Source, Observer, and Fluid, All Moving

Finally, we consider source and observer both moving in a moving fluid in arbitrary directions. The corresponding velocities are denoted by the vectors u, u0, and U.

The simplest way of deriving the expression for the corresponding Doppler shifted frequency is probably to consider the motion from a frame of reference in which the fluid is at rest. The velocities of the source and the observer are then (u− U) and (u₀− U). The Doppler shift depends on the velocity components in the direction of wave travel and we shall denote by ˆk the unit vector for the direction of propagation of the wave from the emission point to the observer. These velocity components are (u−U)· ˆk and (u0−U)· ˆk. The Doppler shifted frequency, by analogy with Eq. 3.14, is then

f^= f [1 − (m0− M) · ˆk]/[1 − (m − M) · ˆk], (3.15) where m= u0.c, m0= u0/c, and M= U/c.

3.1.7 Problems